# Homework Help: Integral in the form 1/sqrt

1. Jun 21, 2007

### sokratesla

integral in the form dx/sqrt(x^2 + a^2)

1. The problem statement, all variables and given/known data
# Hi, this is my first post!
# I have to solve an integral in my electrodynamics study for evaluation of an electric potential. They are in the form $$V(\mathbf{r})=\frac{1}{4\pi\epsilon_0}\int \rho(\mathbf{r'})/\abs(\mathbf{r}-\mathbf{r'}) d\mathbf{r'}$$ In order to find the potential on the top of a conical surface I have to solve an integral in the following form:

2. Relevant equations
$$\int \frac{dx}{\sqrt{x^2+a^2}}$$

3. The attempt at a solution
I solved another problem of the kind
$$\int \frac{x dx}{\sqrt{x^2+a^2}}$$
but stucked with this one.
Can you give a hint, or help in order to solve this integral.

http://integrals.wolfram.com/index.jsp
Mathematica gives $$\log(x+\sqrt{x^2+a^2})$$

Last edited: Jun 21, 2007
2. Jun 21, 2007

### malawi_glenn

Okay, the attempt at a solution?

Do you own a handbook on mathematical identites, have you search the internet and/or your books on one-dimensional calculus?

3. Jun 21, 2007

### bel

Have you tried any trigonometric or hyperbolic function substutions? Might work out to be arcosh or arsinh of some not so pretty thing, my guess.

Let's see, int(1/sqrt(x^2+a^2))dx = arsinh (x/a) , a>0

Last edited: Jun 21, 2007
4. Jun 21, 2007

### malawi_glenn

yes that is correct; $$\ln(x+\sqrt{x^2+a^2})$$
it is.

5. Jun 21, 2007

### sokratesla

# Thanks for the reply!
# Here is my attempt. I found this substitution in my Calculus Book and thought it may be useful.
$$x\equiv a\tan(\theta)$$
so the denominator becomes:
$$a\sqrt{\tan^2(\theta)+1}$$ and integral becomes $$\frac{\cos{\theta}dx}{a}$$
$$dx=a\sec^2(\theta)$$
and at last the integral after substitution is:
$$\int \frac{d\theta}{\cos{\theta}}$$

Last edited: Jun 21, 2007
6. Jun 21, 2007

### malawi_glenn

Okay, now I am confused what you want to calculate.

$$V(\mathbf{r})=\frac{1}{4\pi\epsilon_0}\int ..$$

or

$$\int \frac{dx}{\sqrt{x^2+a^2}}$$

??

What is the form of the charge distribution?

7. Jun 21, 2007

### sokratesla

# Interesting, Maxima gives this as an answer. Mathematica and Maxima seems to give different answers but, then, may be this expressions are equal.

8. Jun 21, 2007

### sokratesla

# Second one. First one is the general formula for vector potentials. But I reduced my problem to the second integral.

9. Jun 21, 2007

### malawi_glenn

Okay, then:

$$\int \frac{dx}{\sqrt{x^2+a^2}} = \ln(x+\sqrt{x^2+a^2})$$

10. Jun 21, 2007

### VietDao29

Yes, whenever you see something like x2 + a2, the first thing you'd like to try would be the substitution: x = atan(t).

After arriving at:
$$\int \frac{d (\theta)}{\cos \theta}$$, we notice that the power of the cosine function is 1, ie, odd, so, we'll use the u-substitution: $$u = \sin \theta$$, like this:
$$\int \frac{d (\theta)}{\cos \theta} = \int \frac{\cos \theta d (\theta)}{\cos ^ 2 \theta} = \int \frac{\cos \theta d (\theta)}{1 - \sin ^ 2 \theta} = \int \frac{du}{1 - u ^ 2}$$.
Hopefully, you can go from here, right?

----------------------------

Or, a different way to tackle this integral is to use Euler Substitution.
Let $$t = \sqrt{x ^ 2 + a ^ 2} + x$$
Take the differential of both sides yields:
$$dt = \frac{x}{\sqrt{x ^ 2 + a ^ 2}} dx + dx = \left(\frac{x}{\sqrt{x ^ 2 + a ^ 2}} + 1 \right) dx = \frac{x + \sqrt{x ^ 2 + a ^ 2}}{\sqrt{x ^ 2 + a ^ 2}} = \frac{t}{\sqrt{x ^ 2 + a ^ 2}} dx$$

Isolate x's, and t's to one side, we have:
$$\Rightarrow \frac{dt}{t} = \frac{dx}{\sqrt{x ^ 2 + a ^ 2}}$$
Now, integrate both sides gives:
$$\int \frac{dx}{\sqrt{x ^ 2 + a ^ 2}} = \int \frac{dt}{t} = \ln|t| + C = \ln|x + \sqrt{x ^ 2 + a ^ 2}| + C$$

----------------------------

Or, to make it more convenient, you can try to learn this integral by heart, it's pretty common.

$$\int \frac{dx}{\sqrt{x^2+a^2}} = \ln(x+\sqrt{x^2+a^2}) + C$$

Last edited: Jun 21, 2007
11. Jun 21, 2007

### George Jones

Staff Emeritus
y = arcsinh(x/a)

sinhy = x/a

(e^y - e^-y)/2 = x/a

Multiply by e^y and rearrange.

e^(2y) - 2x/a e^y - 1 = 0

Use the formula for the roots of a quadratic and simplify.

e^y = (x +/- sqrt(x^2 + a^2))/a

y = ln(x + sqrt(x^2 + a^2) - lna

So, up to a constant of integration, the two answers are the same.

12. Jun 21, 2007

### sokratesla

# Yes, it is the answers. But everywhere what I found is the answer. But I want to learn the intermediate steps too. Otherway I'll feel like I'm cheating in the exam if I memorize the table without learning its proof.
http://www.sosmath.com/tables/integral/integ11/integ11.html
# Here there are two answers.

# I tried the substitution x=a*tan(u) and changed the integral to du / cos(u) but stopped there.

13. Jun 21, 2007

### malawi_glenn

hehe yeah, but is it an exam on integral calculus or electrodynamics?..
Are you not allowed to use collection of formulas and so on?

14. Jun 21, 2007

### bel

let x = a sinh (psi)
dx = a cosh (psi) d(psi)

then int(1/sqrt(x^2+a^2))dx = int (acosh(psi)/ sqrt((a^2)(sinh(psi))^2 + a^2)) d(psi)

= int (a cosh (psi)/ a cosh (psi) ) d(psi)
= int (1) d(psi)
= psi+C

but, x = a sinh (psi)
sihn (psi)= (x/a)
(psi) = arsinh (x/a)

so, int(1/sqrt(x^2+a^2))dx = psi+C = arsinh (x/a) +C

my guess is that ln(x+ sqrt (x^2+a^2)) and this differs by a constant.

sorry I don't know how to type maths symbols, or this would have looked nicer.

15. Jun 21, 2007

### malawi_glenn

16. Jun 21, 2007

### bel

thanks, malawi_glenn. =)

but before I learn all that, the ln(u) thing probably came from the fact that sinh(u)+cosh(u)=e^(u).

17. Jun 21, 2007

### sokratesla

# Thanks for these important advises. du/(1-u^2) seems much simpler. I think I can solve it looking my Calculus book.

# And many thanks to everybody! Have a nice day!

18. Jun 21, 2007

### sokratesla

# It's and electrodynamics exam. In general, the exams are openbook or instructors give some important formulas. But they are all physics books and formulas.
# And, I think, it's a shame for last year physics students, like me, not able to solve basic integrals.:shy:
# Thanks for your help.

19. Jun 21, 2007

### bob1182006

Actually this is true for
$$\int \frac{dx}{\sqrt{x^2 \pm a^2}}=\ln(x+\sqrt{x^2 \pm a^2})$$